Is it possible to find an analytic form for the limit of the infinite product:

$$ \prod_{n=1}^\infty\frac{1+x^{\delta^n}}{2} $$

where $ x>0 $ and $0<\delta<1$?

  • 4
    $\begingroup$ You may start with $\,\delta=\dfrac 12\,$ : I obtained $\dfrac{x-1}{\log(x)}$. $\endgroup$ – Raymond Manzoni Aug 27 '14 at 16:19
  • $\begingroup$ Thanks! Can you give me hint of how you got there? $\endgroup$ – Edmund Crawley Aug 28 '14 at 17:25
  • 1
    $\begingroup$ For $\,\delta=\dfrac 12\,$ (the other cases should not be so easy...) you only need to expand the product of the $m$ first numerators in increasing powers of $x$. You should get a geometric series and finally the fraction $$P_m(x):=\frac {1-x}{(1-x^{1/2^m})2^m}$$ At the limit $m\to +\infty$ you'll get $\dfrac{x-1}{\log(x)}$. I don't know how to handle the general case $\,\delta\neq\dfrac 12$. For $\,\delta =\dfrac 1d$ only the powers represented in base $d$ with $0$ and $1$ will be considered (I didn't search further...) $\endgroup$ – Raymond Manzoni Aug 28 '14 at 19:42

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